On the Growth of Meromorphic Solutions of Certain Nonlinear Difference Equations

Abstract

By Cartan’s version of Nevanlinna’s theory, we prove the following result: let m and n be two positive integers satisfying \(n\ge 2+m,\) let \(p\not \equiv 0\) be a polynomial, let \(\eta \ne 0\) be a finite complex number, let \(\omega _{1}, \omega _{2}, \ldots , \omega _{m}\) be m distinct finite nonzero complex numbers, and let \(H_{j}\) be either exponential polynomials of degree less than q,  or an ordinary polynomial in z for \(0\le j\le m\), such that \(H_{j}\not \equiv 0\) for \(1\le j\le m.\) Suppose that \(f\not \equiv \infty \) is a meromorphic solution of the difference equation:

$$\begin{aligned} f^n(z)+p(z)f(z+\eta )&=H_0(z)+H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}\\&\quad +\cdots +H_m(z)e^{\omega _{m}z^{q}}, \end{aligned}$$

such that the hyper-order of f satisfies \(\rho _2(f)<1.\) Then, f reduces to a transcendental entire function, such that either \(n=m+2\) with \(H_0\not \equiv 0\) and \(\lambda (f)=\rho (f)=q,\) or \(m=2,\) \(H_0=0\) and:

$$\begin{aligned} f(z)=\frac{H_1(z-\eta )e^{\omega _{1}(z-\eta )^{q}}}{p(z-\eta )} \end{aligned}$$

with

$$\begin{aligned} H^n_1(z)=p^n(z)H_2(z+\eta )e^{\omega _2P_{q-1}(z)}\quad \text {and}\quad P_{q-1}(z)=\sum \limits _{k=1}^q\left( {\begin{array}{c}q\\ k\end{array}}\right) \eta ^kz^{q-k}. \end{aligned}$$

This result improves Theorems 1.1 and 1.3 from [19] by removing some assumptions of theirs. An example is provided to show that some results obtained in this paper, in a sense, are the best possible.

Introduction and Main Results

In recent years, the difference variant of the Nevanlinna theory has been established in [5, 8] and, in particular, in [7], by Halburd–Korhonen and by Chiang–Feng, independently. Using these theories, some mathematicians in the world began to consider the studies on the growth of meromorphic and entire solutions of certain nonlinear difference equations, and produced many fine works, for example, see [13, 16, 18, 19]. In this paper, we will consider the growth of meromorphic solutions of the following nonlinear difference equation:

$$\begin{aligned} f^n(z)+p(z)f(z+\eta )=H_0(z)+H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}+\cdots +H_m(z)e^{\omega _{m}z^{q}}, \end{aligned}$$
(1.1)

where m and n are the positive integers satisfying \(n\ge m+2,\) and \(p\not \equiv 0\) is a polynomial, while \(H_j\) is either an exponential polynomial of degree less than q or an ordinary polynomial in z for \(0\le j\le m,\) while \(\eta \ne 0\) is a finite complex value, and \(\omega _{1}, \omega _{2}, \ldots , \omega _{m}\) are m distinct finite nonzero complex numbers. Here, the notion of an exponential polynomial is defined as follows (cf. [11]): an exponential polynomial of order \(q\ge 1,\) which is an entire function of the form:

$$\begin{aligned} g(z)= P_1(z)e^{Q_1(z)} + \cdots + P_k(z)e^{Q_k(z)}, \end{aligned}$$

where \(P_j\) and \(Q_j\) are polynomials in z for \(1\le j\le k\), such that \(\max \nolimits _{1\le j\le k}\{\deg (Q_j)\} = q\) (cf. [14]). Following Steinmetz [14], such a function can be written in the form:

$$\begin{aligned} g(z) = H_1(z)e^{\omega _1z^q}+ \cdots + H_m(z)e^{\omega _mz^q}, \end{aligned}$$

where \(\omega _1, \ldots , \omega _m (m \le k)\) are pairwise different leading coefficients of the polynomials \(Q_1, Q_1, \ldots \) and \(Q_k,\) and the function \(H_j\) is either an exponential polynomial of order \(\le q-1\) or an ordinary polynomial in z for \(1\le j\le k.\) We use the convention that an exponential polynomial of order zero is an ordinary polynomial.

Throughout this paper, by meromorphic functions, we will always mean meromorphic functions in the complex plane. We described the standard notations of the Nevanlinna theory of meromorphic functions as in [12, 15, 17]. It will be convenient to let E denote any set of positive real numbers of finite linear measure, not necessarily the same at each occurrence. For a nonconstant meromorphic function h, we denote by T(rh) the Nevanlinna characteristic of h and by S(rh) any quantity satisfying \(S(r,h)=o(T(r,h)),\) as r runs to infinity outside of a set of finite logarithmic measure. We say that \(\alpha \) is a small function of f, if \(\alpha \) is a meromorphic function, such that \(T(r,\alpha )=S(r,f)\) (cf. [15]). A polynomial P(zf) is called a differential polynomial in f whenever f is a polynomial in f and its derivatives, with small functions of f as the coefficients. Similarly, a polynomial Q(zf) is called a differential-difference polynomial in f whenever Q(zf) is a polynomial in f,  its derivatives, and its shifts \(f(z+\eta )\) with small functions of f again as the coefficients (cf. [16]). In addition, we denote by \(\rho (f),\) \(\rho _2(f),\) \(\lambda (f)\), and \(\lambda _2(f)\) the order, the hyper-order, the exponent of convergence of zeros, and the hyper-exponent of convergence of zeros of f, respectively. Their definitions can be found in [4, 12]. For convenience, we give their detailed definitions as follows:

Definition 1.1

For a nonconstant meromorphic function f,  the order of f,  the hyper-order of f,  the exponent of convergence of zeros of f,  and the hyper-exponent of convergence of zeros of f are denoted by \(\rho (f),\) \(\rho _2(f),\) \(\lambda (f)\), and \(\lambda _2(f)\), respectively. Their definitions are defined as:

$$\begin{aligned} \rho (f)&=\limsup _{r\rightarrow \infty }\frac{{\log }T(r,f)}{{\log }r},\quad \rho _2(f)=\limsup _{r\rightarrow \infty } \frac{{\log }{\log }T(r,f)}{{\log }r},\\ \lambda (f)&=\limsup _{r\rightarrow \infty } \frac{{\log }N\left( r,\frac{1}{f}\right) }{{\log }r}=\limsup _{r\rightarrow \infty } \frac{{\log }n\left( r,\frac{1}{f}\right) }{{\log }r} \quad \text {and}\\ \lambda _2(f)&=\limsup _{r\rightarrow \infty } \frac{{\log \log }N\left( r,\frac{1}{f}\right) }{{\log }r}=\limsup _{r\rightarrow \infty } \frac{{\log \log }n\left( r,\frac{1}{f}\right) }{{\log }r}\quad \text {respectively}. \end{aligned}$$

We recall the following results due to Yang-Laine [16]:

Theorem 1.2

([16, Theorem 2.4]). Let p and q be polynomials. Then, a nonlinear difference equation:

$$\begin{aligned} f^2(z)+q(z)f(z+1)=p(z) \end{aligned}$$

has no transcendental entire solution of finite order.

Theorem 1.3

([16, Theorem 2.5]). A nonlinear difference equation:

$$\begin{aligned} f^3(z)+q(z)f(z+1)=c\sin (bz), \end{aligned}$$

where q is a nonconstant polynomial and b, \(c\in \mathbb {C}\) are nonzero constants, do not admit entire solutions of finite order. If q is a nonzero constant, then this equation possesses three distinct entire solutions of finite order, provided \(b=3n\pi \) and \(q^3=(-1)^{n+1}\frac{27}{4}c^2\) for a nonzero integer n.

Recently, Zhang and Huang [19] proved the following result which extends Theorems 1.2 and 1.3:

Theorem 1.4

([19, Theorem 1.1]). Let s and n be two positive integers satisfying \(n\ge 2+s,\) let \(p\not \equiv 0\) be a polynomial, let \(\eta \ne 0\) be a finite complex number, let \(\beta _1, \beta _2, \ldots , \beta _s\) be s finite nonzero complex numbers, and let \(\alpha _1, \alpha _2, \ldots , \alpha _s\) be s distinct finite nonzero complex numbers. Suppose that \(\frac{\alpha _i}{\alpha _j}\ne n\) for all \(i, j\in \{1,2,\ldots ,s\},\) and that \(n\alpha _k\ne l_{k_1}\alpha _1+l_{k_2}\alpha _2+\cdots +l_{k_s}\alpha _s\) for \(5\le k\le s\) when \(s\ge 5,\) where \(l_{k_1},l_{k_2},\ldots ,l_{k_s}\in \{0,1,2,\ldots ,n-1\}\) satisfying \(l_{k_1}+l_{k_2}+\cdots +l_{k_s}= n.\) Then, for any nonconstant meromorphic solution f of the equation:

$$\begin{aligned} f^n(z)+p(z)f(z+\eta )=\beta _1(z)e^{\alpha _{1}z}+\beta _2(z)e^{\alpha _{2}z}+\cdots +\beta _s(z)e^{\alpha _{s}z}, \end{aligned}$$
(1.2)

we have \(\rho _2(f)\ge 1.\)

Theorem 1.5

([19, Theorem 1.2]). Let n and s be two positive integers satisfying \(n=s+1,\) let \(p\not \equiv 0\) be a polynomial, let \(\eta \ne 0\) be a finite complex number, let \(\beta _1, \beta _2, \ldots , \beta _s\) be finite nonzero complex numbers, and let \(\alpha _1, \alpha _2, \ldots , \alpha _s\) be s distinct finite nonzero complex numbers. Suppose that \(\frac{\alpha _i}{\alpha _j}\ne n\) for all \(i, j\in \{1,2,\ldots ,s\}.\) Then, any nonconstant meromorphic solution f of Eq. (1.2) of hyper-order \(\rho _2(f)< 1\) must be an entire function, such that \(\rho (f)=\lambda (f)=1.\)

Regarding Theorems 1.4 and 1.5, one may ask the following two questions:

Question 1.6

What can be said about the conclusions of Theorem 1.4, if we remove the assumptions “Suppose that \(\frac{\alpha _i}{\alpha _j}\ne n\) for all \(i, j\in \{1,2,\ldots ,s\},\) and that \(n\alpha _k\ne l_{k_1}\alpha _1+l_{k_2}\alpha _2+\cdots +l_{k_s}\alpha _s\) for \(5\le k\le s\) when \(s\ge 5,\) where \(l_{k_1},l_{k_2},\ldots ,l_{k_s}\in \{0,1,2,\ldots ,n-1\}\) satisfying \(l_{k_1}+l_{k_2}+\cdots +l_{k_s}= n\)” in Theorem 1.4?

Question 1.7

What can be said about the conclusions of Theorem 1.5, if we remove the assumption “Suppose that \(\frac{\alpha _i}{\alpha _j}\ne n\) for all \(i, j\in \{1,2,\ldots ,s\}\)” in Theorem 1.5?

In this direction, we first study the growth of a nonconstant meromorphic solution of Eq. (1.1) which is much more general difference equation than the difference equation (1.2). Indeed, we will prove the following result:

Theorem 1.8

Let m and n be two positive integers satisfying \(n\ge 2+m,\) let \(p\not \equiv 0\) be a polynomial, let \(\eta \ne 0\) be a finite complex number, let \(\omega _{1}, \omega _{2}, \ldots , \omega _{m}\) be m distinct finite nonzero complex numbers, let \(H_{j}\) be either exponential polynomials of degree less than q,  or an ordinary polynomials in z for \(0\le j\le m,\) and let \(H_{j}\not \equiv 0\) for \(1\le j\le m.\) Suppose that \(f\not \equiv \infty \) is a meromorphic solution of the difference equation (1.1), such that the hyper-order of f satisfies \(\rho _2(f)<1.\) Then, f reduces to a transcendental entire function, such that one of the following two cases can occur:

  1. (i)

    The positive integer m in (1.1) satisfies \(m=2,\) and that the entire function \(H_0\) in (1.1) satisfies \(H_0=0.\) Moreover, f can be expressed as:

    $$\begin{aligned} f(z)=\frac{H_1(z-\eta )e^{\omega _{1}(z-\eta )^{q}}}{p(z-\eta )} \end{aligned}$$

    with

    $$\begin{aligned} H^n_1(z)=p^n(z)H_2(z+\eta )e^{\omega _2P_{q-1}(z)}\quad \text {and}\quad P_{q-1}(z)=\sum \limits _{k=1}^q\left( {\begin{array}{c}q\\ k\end{array}}\right) \eta ^kz^{q-k}. \end{aligned}$$
  2. (ii)

    The positive integers m and n in (1.1) satisfy \(n=m+2,\) and the entire function \(H_0\) in (1.1) satisfies \(H_0\not \equiv 0.\) Moreover, the order of f and the exponent of convergence of zeros of f satisfy \(\lambda (f)=\rho (f)=q.\)

By Theorem 1.8, we can get the following two results, which solve Questions 1.6 and 1.7, respectively, and improve Theorems 1.4 and 1.5, respectively:

Corollary 1.9

Let m and n be two positive integers satisfying \(n\ge 2+m\) and \(m\ne 2,\) let \(p\not \equiv 0\) be a polynomial, let \(\eta \ne 0\) be a finite complex number, let \(\omega _{1},\) \(\omega _{2},\) \(\cdots \), \(\omega _{m}\) be m distinct finite nonzero complex numbers, let \(H_{j}\) be either exponential polynomials of degree less than q,  or an ordinary polynomials in z for \(1\le j\le m,\) and let \(H_{j}\not \equiv 0\) for \(1\le j\le m.\) Suppose that \(f\not \equiv \infty \) is a meromorphic solution of the difference equation:

$$\begin{aligned} f^n(z)+p(z)f(z+\eta )=H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}+\cdots +H_m(z)e^{\omega _{m}z^{q}}. \end{aligned}$$
(1.3)

Then, f reduces to a transcendental entire function, such that its hyper-order \(\rho _2(f)\) satisfies \(\rho _2(f)\ge 1.\)

By Theorem 1.8, we also get the following result, which is a supplement of Theorem 1.4 and Corollary 1.9:

Corollary 1.10

Let m and n be two positive integers satisfying \(n\ne 2+m,\) let \(p\not \equiv 0\) be a polynomial, let \(\eta \ne 0\) be a finite complex number, let \(\omega _{1}, \omega _{2}, \ldots , \omega _{m}\) be m distinct finite nonzero complex numbers, let \(H_{j}\) be either exponential polynomials of degree less than q,  or an ordinary polynomials in z for \(0\le j\le m,\) and let \(H_{j}\not \equiv 0\) for \(0\le j\le m.\) Suppose that \(f\not \equiv \infty \) is a meromorphic solution of the difference equation (1.1). Then, f reduces to a transcendental entire function, such that the hyper-order of f satisfies \(\rho _2(f)\ge 1.\)

The following example shows that the assumption of “\(n\ge 2+m\)” in Corollary 1.9, in a sense, is the best possible. The example also shows that the assumption of “\(H_0\not \equiv 0\)” in Corollary 1.10 is necessary.

Example 1.11

Let:

$$\begin{aligned} f(z)=e^{\frac{z}{5}}+e^{-\frac{z}{5}}. \end{aligned}$$
(1.4)

Then, it follows that the entire function f defined as in (1.4) is an entire solution of the difference equation:

$$\begin{aligned} f^5(z)-10f(z+10 \pi i)=5e^{\frac{3z}{5}}+5e^{-\frac{3z}{5}}+e^{z}+e^{-z}, \end{aligned}$$

such that \(\rho (f)=1\) and:

$$\begin{aligned} T(r,f(z))=N\left( r,\frac{1}{f(z)}\right) +O(1)=N\left( r,\frac{1}{e^{\frac{2z}{5}}+1}\right) +O(1)=\frac{2r}{5\pi }\left( 1+o(1)\right) . \end{aligned}$$

Moreover, we can see that \(n=m+1\) and \(n=5.\)

Preliminaries

In this section, we will introduce some results used to prove the main result in the present paper. First of all, we introduce the following result due to Yang and Yi [15].

Lemma 2.1

([15, Theorem 1.62]).   Let \(f_1, f_2,\ldots , f_n\) be nonconstant meromorphic functions, and let \(f_{n+1}\not \equiv 0\) be a meromorphic function such that \(\sum \nolimits _{j=1}^{n+1}f_j=1.\) Suppose that there exists a subset \(\mathrm{I}\subseteq \mathbb {R}^{+}\) with linear measure \(\text {mes} \, \mathrm{I}=\infty \), such that:

$$\begin{aligned} \sum _{i=1}^{n+1} N\left( r,\frac{1}{f_i}\right) +n\sum _{ {\mathop { i\ne j}\limits ^{i=1}}}^{n+1}\overline{N}(r,f_i) <(\sigma + o(1))T(r,f_j),\quad j=1, 2, \ldots , n, \end{aligned}$$

as \(r\in \mathrm{I}\) and \(r\rightarrow \infty ,\) where \(\sigma \) is a real number satisfying \(0\le \sigma <1.\)   Then, \(f_{n+1}=1.\)

For introducing the following lemma, we give the following notation (cf. [6, Definition 2.1]): for a meromorphic function f satisfying \(f\not \equiv 0\) and a positive integer j,  we let \(n_j(r, 1/ f)\) denote the number of zeros of f in \(\{z : |z| \le r\},\) counted in the following manner: a zero of f of multiplicity m is counted exactly k times where \(k = \min \{m, j\}.\) Then, we let \(N_j(r, 1/ f)\) denote the corresponding integrated counting function, that is:

$$\begin{aligned} N_j\left( r,\frac{1}{f}\right) =\int _0^r\frac{n_j\left( t,\frac{1}{f}\right) -n_j\left( 0,\frac{1}{f}\right) }{t}\mathrm{{d}}t+n_j\left( 0,\frac{1}{f}\right) \log r. \end{aligned}$$

Obviously:

$$\begin{aligned} N_1\left( r,\frac{1}{f}\right) =\overline{N}\left( r,\frac{1}{f}\right) ,\quad \overline{N}\left( r,\frac{1}{f}\right) \le N_j\left( r,\frac{1}{f}\right) \le N\left( r,\frac{1}{f}\right) \end{aligned}$$

and

$$\begin{aligned} N_j\left( r,\frac{1}{f}\right) \le j\overline{N}\left( r,\frac{1}{f}\right) . \end{aligned}$$

In 1933, Cartan proved a generalization of the second fundamental theorem, and for certain kinds of questions. Moreover, Cartan’s theorem seems to give the better result than the second fundamental theorem. We now state the simple form of Cartan’s theorem as follows.

Lemma 2.2

(Cantan’s theorem, [3]).   Let \(g_1, g_2,\ldots ,g_p\) be linearly independent entire functions, where \(p\ge 2.\) Suppose that for each complex number z, we have \(\max \{|g_1(z)|, |g_2(z)|,\ldots ,|g_p(z)|\}>0.\) For positive r:

$$\begin{aligned} T(r)=\frac{1}{2\pi }\int _0^{2\pi } u(re^{i\theta })\mathrm{{d}}\theta -u(0), \quad where~u(z)=\sup _{1\le j \le p}\log |g_j(z)|. \end{aligned}$$

Set \(g_{p+1}=g_1+g_2+\cdots +g_p.\) Then, we have:

$$\begin{aligned} T(r)\le \sum _{j=1}^{p+1}N_{p-1}\left( r,\frac{1}{g_j}\right) +S(r)\le (p-1)\sum _{j=1}^{p+1}\overline{N}\left( r,\frac{1}{g_j}\right) +S(r), \end{aligned}$$

where S(r) is a quantity satisfying:

$$\begin{aligned} S(r)=O(\log T(r))+O(\log r), \quad \text {as}~r\rightarrow \infty ~\text {and}~r\notin E. \end{aligned}$$

If at least one of the quotients \(g_j/g_m\) is a transcendental function, then:

$$\begin{aligned} S(r)=o(T(r))\quad \text {as}~r\rightarrow \infty ~\text {and}~r\notin E, \end{aligned}$$

while if all the quotients \(g_j/g_m\) are rational functions, then:

$$\begin{aligned} S(r)\le -\frac{1}{2}p(p-1)\log r+O(1), \quad \text {as}~r\rightarrow \infty ~\text {and}~r\notin E. \end{aligned}$$

Here, \(E\subset (0,+\infty )\) is a subset of finite linear measure.

We also need the following result:

Lemma 2.3

[3] Assume that the hypotheses of Lemma 2.2 hold. Then, for any j and m,  we have:

$$\begin{aligned} T\left( r,\frac{g_j}{g_m}\right) \le T(r)+O(1), \quad \text {as}~r\rightarrow \infty , \end{aligned}$$

and for any j,  we have:

$$\begin{aligned} N\left( r,\frac{1}{g_j}\right) \le T(r)+O(1), \quad \text {as} ~r\rightarrow \infty . \end{aligned}$$

For introducing the following two lemmas, we introduce the notion of the convex hull of a subset of the complex plane \(\mathbb {C}\) (cf. [14]): the convex hull of a subset \(W\subset \mathbb {C},\) denoted as \(\mathrm{co(W)},\) is the intersection of all convex sets containing the set W. If W contains only finitely many elements, then \(\mathrm{co(W)}\) is obtained as an intersection of finitely many closed half-planes, and hence, \(\mathrm{co(W)}\) is either a compact polygon (with a nonempty interior) or a line segment. We denote the perimeter of \(\mathrm{co(W)}\) by \(\mathrm{C(co(W))}.\) If \(\mathrm{co(W)}\) is a line segment, then \(\mathrm{C(co(W))}\) equals to twice the length of this line segment. Throughout the rest of the paper, we fix the notations for \(W = \{ \overline{\omega }_1, \overline{\omega }_2, \ldots ,\overline{\omega }_m\}\) and \(W_0 = \{0, \overline{\omega }_1, \overline{\omega }_2,\ldots , \overline{\omega }_m\}.\)

Lemma 2.4

([14, Satz 1 and Satz 2]). Let f be given as:

$$\begin{aligned} f(z)=H_0(z)+H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}+\cdots +H_m(z)e^{\omega _{m}z^{q}}, \end{aligned}$$

where m and q are positive integers, \(H_j\) is either an exponential polynomial of degree less than q or an ordinary polynomial in z for \(0\le j\le m,\) and that \(\omega _{1}, \omega _{2}, \ldots , \omega _{m}\) are m distinct finite nonzero complex numbers. Suppose that \(H_j\not \equiv 0\) for \(1\le j\le m.\) Then:

$$\begin{aligned} T\left( r,f\right) =C(co(W_0))\frac{r^q}{2\pi }+o(r^q). \end{aligned}$$

If \(H_0\not \equiv 0,\) then:

$$\begin{aligned} m\left( r,\frac{1}{f}\right) =o(r^q). \end{aligned}$$

If \(H_0= 0,\) then:

$$\begin{aligned} N\left( r,\frac{1}{f}\right) =C(co(W_0))\frac{r^q}{2\pi }+o(r^q). \end{aligned}$$

The following result is due to Halburd et al.  [9]:

Lemma 2.5

([9, Lemma 8.3]). Let \(T(r):[0,+\infty )\rightarrow [0,+\infty )\) be a nondecreasing continuous function and let \(s\in \mathbb {R}^+\). If the hyper-order of T is strictly less than one, i.e., \(\limsup \nolimits _{r\rightarrow \infty }\frac{\log \log T(r)}{\log r}=\zeta <1,\) and let \(\delta \in (0,1-\zeta )\), then \(T(r+s)=T(r)+o\left( \frac{T(r)}{r^\delta }\right) ,\) where r runs to infinity outside of a set of finite logarithmic measure.

The following result is proved by Chiang and Feng [5]:

Lemma 2.6

([5, Theorem 2.1]).   Let f be a meromorphic function of order \(\rho (f)=\rho < \infty ,\) and let \(\eta \) be a fixed nonzero complex number, and then, for each \(\varepsilon >0,\) we have:

$$\begin{aligned} T(r,f(z +\eta )) = T (r,f(z))+ O(r^{\rho -1+\varepsilon })+O(\log r). \end{aligned}$$

Proof of Theorem 1.8

First, for convenience, we give the following notation of some assumptions of Theorem 1.8:

(H). Suppose that \(H_0\) and \(H_{j}\not \equiv 0\) are either exponential polynomials of degree less than q,  or the ordinary polynomials in z for \(1\le j\le m,\) and that \(\omega _{1}, \omega _{2},\ldots ,\omega _{m}\) are m distinct finite nonzero complex numbers.

By noting that \(f\not \equiv \infty \) is a meromorphic solution of (1.1), we consider the following two cases.

Case 1.   Suppose that f is a constant, say \(f=c,\) where c is some finite constant. Then, by rewriting (1.1), we have:

$$\begin{aligned} H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}+\cdots +H_m(z)e^{\omega _{m}z^{q}}=c^n+cp(z)-H_0(z). \end{aligned}$$
(3.1)

Clearly, the left hand of (3.1) is an exponential polynomial of order q and the right hand of (3.1) is an exponential polynomial of order less than q,  or an ordinary polynomial in z , this is a contradiction. Thus, \(c^n + cp-H_0 = 0.\) This together with (3.1) gives:

$$\begin{aligned} H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}+\cdots +H_m(z)e^{\omega _{m}z^{q}}=0. \end{aligned}$$
(3.2)

By dividing \(e^{\omega _{1}z^{q}}\) on two sides of (3.2), we have:

$$\begin{aligned} H_2(z)e^{(\omega _{2}-\omega _{1})z^{q}}+\cdots +H_m(z)e^{(\omega _{m}-\omega _{1})z^{q}}=-H_1(z). \end{aligned}$$
(3.3)

In the same manner as the above, we deduce \(-H_1=0,\) which contradicts the assumption \(H_1\not \equiv 0.\)

Case 2.   Suppose that f is a nonconstant meromorphic function such that \(\rho _2(f) < 1.\) First of all, we prove that f is a transcendental entire function. Indeed, we suppose that \(z_0\in \mathbb {C}\) is a pole of f(z) of multiplicity equal to \(\nu .\) Then, it follows by (1.1) and the assumption of Theorem 1.8 that \(z_0+\eta \) is a pole of f(z) of multiplicity equal to \(n\nu \) at least. By substituting \(z=z_0+\eta \) into (1.1), we have:

$$\begin{aligned} f^n(z_0+\eta )+p(z_0+\eta )f(z+2\eta )=H_0(z_0+\eta )+\sum \limits _{j=1}^m H_j(z_0+\eta )e^{\omega _{j}(z_0+\eta )^{q}}, \end{aligned}$$

which implies that \(z_0+2\eta \) is a pole of f(z) of multiplicity equal to \(n^2\nu \) at least. Following this step, we can find a sequence \(\{z_0+j\eta \}_{j=0}^{\infty }\) of the poles of f(z). Here, \(z_0+j\eta \) is a pole of f(z) of multiplicity equal to \(n^j\nu \) for each positive integer j. Therefore, for each positive integer j,  we have:

$$\begin{aligned} n(j |\eta |+\left| z_0 \right| +1,f) \ge \nu + n\nu +\cdots + n^j\nu , \end{aligned}$$

which together with the assumption \(n\ge m+2\ge 3\) gives:

$$\begin{aligned} \lambda _2\left( \frac{1}{f}\right)= & {} \limsup _{r\rightarrow \infty } \frac{{\log }{\log }n(r,f)}{{\log }r}\ge \limsup _{j\rightarrow \infty } \frac{{\log }{\log }n(j|\eta |+|z_0|+1,f(z))}{\log (j|\eta |+|z_0|+1)}\nonumber \\\ge & {} \limsup _{j\rightarrow \infty }\frac{{\log }{\log }n^j}{\log (j|\eta |+|z_0|+1)}=\limsup _{j\rightarrow \infty }\frac{\log j+\log \log n}{\log (j|\eta |+|z_0|+1)}=1.\nonumber \\ \end{aligned}$$
(3.4)

On the other hand, by Definition 1.1, we have \(\rho _2(f)\ge \lambda _2\left( \frac{1}{f}\right) .\) This together with (3.4) gives \(\rho _2(f)\ge 1,\) which contradicts the assumption \(\rho _2(f)<1.\) Therefore, f is a nonconstant entire function of hyper-order \(\rho _2(f)<1.\) We consider the following two subcases:

Subcase 2.1   Suppose that f is a nonconstant polynomial. Then:

$$\begin{aligned} T(r,f)=O(\log r). \end{aligned}$$
(3.5)

By rewriting (1.1), we have:

$$\begin{aligned} H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}+\cdots +H_m(z)e^{\omega _{m}z^{q}}=f^n(z)+p(z)f(z+\eta )-H_0(z). \end{aligned}$$
(3.6)

By (3.5) and the assumption (H), we deduce for \(1\le j\le m\) and \(1\le k\le m\) that:

$$\begin{aligned} T(r, H_j(z))+T(r,f^n(z)+p(z)f(z+\eta )-H_0(z))=o(T(r, e^{\omega _{k}z^{q}}), \end{aligned}$$
(3.7)

as \(r\rightarrow \infty .\)

By (3.6), (3.7) and Lemma 2.1, we deduce (3.2) and:

$$\begin{aligned} f^n(z)+p(z)f(z+\eta )-H_0(z)=0. \end{aligned}$$
(3.8)

By (3.8), we see that (3.6) can be rewritten as (3.2). By dividing \(e^{\omega _{1}z^{q}}\) on two sides of (3.2), we have (3.3). By (3.3), (3.7), Lemma 2.1, the assumption (H) and Hayman [10, p.7], we deduce \(H_1=0,\) which is impossible.

Subcase 2.2   Suppose that f is a transcendental entire function. We consider the following two subcases.

Subcase 2.2.1.   Suppose that \(H_0=0.\) Then, (1.1) can be rewritten as:

$$\begin{aligned} f^n(z)=H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}+\cdots +H_m(z)e^{\omega _{m}z^{q}}-p(z)f(z+\eta ). \end{aligned}$$
(3.9)

By the assumption (H), we deduce that \(H_1(z)e^{\omega _{1}z^{q}}, H_2(z)e^{\omega _{2}z^{q}}, \ldots , H_m(z)e^{\omega _{m}z^{q}}\) are m distinct linearly independent entire functions. Combining this with (3.9), we consider the following two subcases:

Subcase 2.2.1.1.   Suppose that \(H_1(z)e^{\omega _{1}z^{q}}, H_2(z)e^{\omega _{2}z^{q}}, \ldots , H_m(z)e^{\omega _{m}z^{q}},\) \(-p(z)f(z+\eta )\) are \(m+1\) linearly dependent entire functions in the complex plane. Then, \(p(z)f(z+\eta )\) can be linearly expressed by the m distinct entire functions \(H_1(z)e^{\omega _{1}z^{q}}, H_2(z)e^{\omega _{2}z^{q}}, \ldots , H_m(z)e^{\omega _{m}z^{q}}.\) This means that there exist m finite complex constants \(k_1, k_2, \ldots , k_m\), such that:

$$\begin{aligned} p(z)f(z+\eta )=\sum \limits _{j=1}^m k_jH_j(z)e^{\omega _{j}z^{q}}. \end{aligned}$$
(3.10)

By substituting (3.10) into (3.9), we deduce:

$$\begin{aligned} f^n(z)=\sum \limits _{j=1}^m (1-k_j)H_j(z)e^{\omega _{j}z^{q}}. \end{aligned}$$
(3.11)

Without loss of generality, we suppose that \(1-k_j\ne 0\) for \(1\le j\le m.\) Moreover, by the assumption (H), we deduce that the m distinct entire functions:

$$\begin{aligned} (1-k_1)H_1(z)e^{\omega _{1}z^{q}}, (1-k_2)H_2(z)e^{\omega _{2}z^{q}}, \ldots , (1-k_m)H_m(z)e^{\omega _{m}z^{q}} \end{aligned}$$
(3.12)

in (3.11) are linearly independent transcendental entire functions.

Suppose that \( f^n(z)\) and the m distinct entire functions in (3.12) have infinitely many common zeros \(a_{1,1}, a_{1,2}, \ldots ,\) \(a_{1,k}, \ldots ,\) in the complex plane, such that \(a_{1,k}\rightarrow \infty .\) Then, by Weierstrass’s factorization theorem (cf. [2, Theorem 7 in Chapter 5, p.195]), we can find that there exists an entire function \(\gamma _1(z)\) with the common zeros of \( f^n(z)\) and the m distinct entire functions in (3.12) for zeros, where each such common zero \(a_{1,k}\in \{a_{1,k}\}_{k=1}^{\infty }\) is counted according to the minimum of all the multiplicities of \(a_{1,k}\) as the zero of \( f^n(z)\) and the m distinct entire functions in (3.12), respectively, such that:

$$\begin{aligned} \gamma _1(z)=z^{m_{1,0}}e^{g(z)}\prod \limits _{k=1}^{\infty }\left( 1-\frac{z}{a_{1,k}}\right) e^{\frac{z}{a_{1,k}}+\frac{1}{2}\left( \frac{z}{a_{1,k}}\right) ^2 +\cdots +\frac{1}{m_{1,k}}\left( \frac{z}{a_{1,k}}\right) ^{m_{1,k}}}, \end{aligned}$$
(3.13)

where the product is taken over all \(a_{1,k}\ne 0.\) Here, \(m_{1,0}, m_{1,1}, \ldots ,\) \(m_{1,k}, \ldots ,\) are certain nonnegative integers, and g(z) is a constant or a nonconstant polynomial of degree less than or equal to \(q-1.\) Combining this with the assumption (H), we deduce:

$$\begin{aligned} N\left( r,\frac{1}{\gamma _1}\right) \le \sum \limits _{j=1}^m N\left( r,\frac{1}{H_j}\right) \le \sum \limits _{j=1}^m T\left( r,H_j \right) +O(1)\le O(r^{q-1}). \end{aligned}$$
(3.14)

By (3.14) and Ash [1, Theorem 4.3.6], we have:

$$\begin{aligned} \lambda (\gamma _1)=\rho (\gamma _1)\le q-1. \end{aligned}$$
(3.15)

By dividing \(\gamma _1\) defined as in (3.13) on two sides of (3.11), we have:

$$\begin{aligned} \frac{f^n(z)}{\gamma _1(z)}=\sum \limits _{j=1}^m (1-k_j)H_{1,j}(z)e^{\omega _{j}z^{q}}, \end{aligned}$$
(3.16)

where:

$$\begin{aligned} H_{1,j}(z)=\frac{H_j(z)}{\gamma _1(z)}, \quad 1\le j\le m. \end{aligned}$$
(3.17)

Then, by (3.17) and the obtained result that the m distinct transcendental entire functions in (3.12) are linearly independent in the complex plane, we deduce that the m entire functions \((1-k_1)H_{1,1}(z)e^{\omega _{1}z^{q}}, (1-k_2)H_{1,2}(z)e^{\omega _{2}z^{q}}, \ldots , (1-k_m)H_{1,m}(z)e^{\omega _{m}z^{q}}\) on the right-hand side of (3.16) are still linearly independent in the complex plane. Moreover, by (3.16) and the definition of the entire function \(\gamma _1\) in (3.13), we can deduce that the entire function \(\frac{f^n(z)}{\gamma _1(z)}\) and the m entire functions \((1-k_1)H_{1,1}(z)e^{\omega _{1}z^{q}}, (1-k_2)H_{1,2}(z)e^{\omega _{2}z^{q}}, \ldots , (1-k_m)H_{1,m}(z)e^{\omega _{m}z^{q}}\) on the right-hand side of (3.16) have no common zeros in the complex plane. Furthermore, by (3.15), (3.17), the assumption (H), and Hayman [10, p.7], we deduce:

$$\begin{aligned} \frac{H_{1,i}(z)}{H_{1,j}(z)}e^{(\omega _{i}-\omega _{j})z^{q}}=\frac{H_i(z)}{H_j(z)}e^{(\omega _{i}-\omega _{j})z^{q}} \end{aligned}$$
(3.18)

are transcendental meromorphic functions for \(1\le i<j<m.\) For convenience, now, we set:

$$\begin{aligned} \frac{f^n(z)}{\gamma _1(z)}=g_{1,m+1}(z) \end{aligned}$$
(3.19)

and

$$\begin{aligned} (1-k_j)H_{1,j}(z)e^{\omega _{j}z^{q}}=g_{1,j}(z) \quad \text {for} \quad 1\le j\le m. \end{aligned}$$
(3.20)

Then, it follows by (3.19), (3.20) and the above analysis that (3.16) can be rewritten as:

$$\begin{aligned} g_{1,m+1}=\sum \limits _{j=1}^m g_{1,j} \end{aligned}$$
(3.21)

such that \(\max \nolimits _{1\le j\le m}\{|g_{1,j}(z)|\}>0\) for each \(z\in \mathbb {C},\) and \(g_{1,1}, g_{1,2}, \ldots , g_{1,m}\) in (3.21) are m linearly independent transcendental entire functions. Therefore, by (3.15), (3.17), (3.18)–(3.21) and Lemmas 2.2 and 2.3, we deduce:

$$\begin{aligned} nN\left( r,\frac{1}{f}\right)= & {} N\left( r,\frac{1}{g_{1,m+1}}\right) +O(r^{q-1})\le T_1(r)+O(r^{q-1})\nonumber \\\le & {} \sum _{k=1}^{m+1}N_{m-1}\left( r,\frac{1}{g_{1,k}}\right) +O(r^{q-1})+S_1(r)\nonumber \\\le & {} (m-1)\overline{N}\left( r,\frac{1}{f}\right) +O(r^{q-1})+S_1(r)\nonumber \\\le & {} (m-1)N\left( r,\frac{1}{f}\right) +O(r^{q-1})+S_1(r), \end{aligned}$$
(3.22)

where

$$\begin{aligned} T_1(r)=\frac{1}{2\pi }\int _0^{2\pi } u_1(re^{i\theta })\mathrm{{d}}\theta -u_1(0)\quad \text {with} \quad u_1(z)=\sup _{1\le k\le m}\{\log |g_{1,k}(z)|\}, \end{aligned}$$

and \(S_1(r)\) is a quantity satisfying::

$$\begin{aligned} S_1(r)=O(\log T_1(r))+O(\log r), \end{aligned}$$
(3.23)

as \(r\rightarrow \infty \) and \(r\notin E.\) Here, \(E\subset (0,+\infty )\) is a subset of finite linear measure.

By (3.22) and the assumption \(n\ge m+2\), we deduce:

$$\begin{aligned} N\left( r,\frac{1}{f}\right) =O(r^{q-1})+S_1(r) \end{aligned}$$
(3.24)

and

$$\begin{aligned} T_1(r)\le (m-1)N\left( r,\frac{1}{f}\right) +O(r^{q-1})+S_1(r)\le (m-1)T\left( r,f\right) +O(r^{q-1})+S_1(r), \end{aligned}$$
(3.25)

as \(r\not \in E\) and \(r\rightarrow +\infty .\) Since f is a transcendental entire function, by (3.23) and (3.25), we deduce:

$$\begin{aligned} S_1(r)=o(T(r,f))+O(r^{q-1}), \end{aligned}$$
(3.26)

as \(r\not \in E\) and \(r\rightarrow +\infty .\)

By (3.24) and (3.26), we deduce:

$$\begin{aligned} N\left( r,\frac{\gamma _1(z)e^{\omega _{1}z^{q}}}{f^n(z)}\right) \le nN\left( r,\frac{1}{f(z)}\right) \le O(r^{q-1})+o(T(r,f(z)))=o(r^q), \end{aligned}$$
(3.27)

as \(r\not \in E\) and \(r\rightarrow +\infty .\)

On the other hand, by dividing \(e^{\omega _{1}z^{q}}\) on two sides of (3.16), we have:

$$\begin{aligned} \frac{f^n(z)}{\gamma _1(z)e^{\omega _{1}z^{q}}}=(1-k_1)H_{1,1}(z)+\sum \limits _{j=2}^m (1-k_j)H_{1,j}(z)e^{(\omega _{j}-\omega _{1})z^{q}}. \end{aligned}$$
(3.28)

By the assumption (H) and the supposition \((1-k_j)H_{1,j}\not \equiv 0\) for \(1\le j\le m,\) we deduce by (3.15), (3.17), (3.24)–(3.28), Lemma 2.4, and the first fundamental theorem that:

$$\begin{aligned} m\left( r,\frac{\gamma _1(z)e^{\omega _{1}z^{q}}}{f^n(z)}\right)&=m\left( r,\frac{1}{(1-k_1)H_{1,1}(z)+\sum \nolimits _{j=2}^m (1-k_j)H_{1,j}(z)e^{(\omega _{j}-\omega _{1})z^{q}}}\right) \\&=o(r^q) \end{aligned}$$

and

$$\begin{aligned} N\left( r,\frac{\gamma _1(z)e^{\omega _{1}z^{q}}}{f^n(z)}\right)= & {} N\left( r,\frac{1}{(1-k_1)H_{1,1}(z)+\sum \nolimits _{j=2}^m (1-k_j)H_{1,j}(z)e^{(\omega _{j}-\omega _{1})z^{q}}}\right) \nonumber \\= & {} T\left( r,\frac{1}{(1-k_1)H_{1,1}(z)+\sum \nolimits _{j=2}^m (1-k_j)H_{1,j}(z)e^{(\omega _{j}-\omega _{1})z^{q}}}\right) \nonumber \\&+o(r^q)\nonumber \\= & {} T\left( r,\frac{\gamma _1(z)e^{\omega _{1}z^{q}}}{f^n(z)}\right) +o(r^q)=C(co(\hat{W}_0))\frac{r^q}{2\pi }+o(r^q), \end{aligned}$$
(3.29)

where \(\hat{W}_0 = \{0, \overline{\omega _2-\omega _1}, \overline{\omega _3-\omega _1}, \ldots , \overline{\omega _m-\omega _1}\}.\)

By (3.29) and (3.27), we get a contradiction.

Suppose that \( f^n(z)\) and the m distinct entire functions in (3.12) have at most finitely many common zeros in the complex plane, say that \(a_{1,1}, a_{1,2}, \ldots , a_{1,l_1}\) are all the distinct common nonzero zeros of \( f^n(z)\) and the m distinct transcendental entire functions of (3.12) in the complex plane, where \(l_1\) is a positive integer, and that each such common zero \(a_{1,j}\in \{a_{1,j}\}_{j=1}^{l_1}\) is counted according to the minimum positive integer \(\tilde{m}_{1,j}\) of all the multiplicities of \(a_{1,j}\) as the zero of \( f^n(z)\) and the m distinct transcendental entire functions of (3.12), respectively. Now, we let:

$$\begin{aligned} P_{1,1}(z)=z^{m_{1,0}}\prod \limits _{j=1}^{l_1}\left( z-a_{1,j}\right) ^{\tilde{m}_{1,j}}, \end{aligned}$$
(3.30)

where \(m_{1,0}\) is the nonnegative integer that is defined as in (3.13). Next, we replace the polynomial \(P_{1,1}\) defined as in (3.30) with the entire function \(\gamma _1\) defined as in (3.13), and then use the lines as the above, we can get a contradiction.

Subcase 2.2.1.2.   Suppose that:

$$\begin{aligned} H_1(z)e^{\omega _{1}z^{q}}, \quad H_2(z)e^{\omega _{2}z^{q}}, \quad \ldots , H_m(z)e^{\omega _{m}z^{q}}, -p(z)f(z+\eta ) \end{aligned}$$
(3.31)

are \(m+1\) linearly independent entire functions in the complex plane.

Suppose that \( f^n(z)\) and the \(m+1\) distinct entire functions in (3.31) have infinitely many common zeros \(a_{2,1}, a_{2,2}, \ldots , a_{2,k}, \ldots ,\) in the complex plane such that \(a_{2,k}\rightarrow \infty .\) Then, by Weierstrass’s factorization theorem (cf. [2, Theorem 7 in Chapter 5, p.195]), we can find that there exists an entire function \(\gamma _2(z)\) with the common zeros of \( f^n(z)\) and the \(m+1\) distinct entire functions in (3.31) for zeros, where each such common zero \(a_{2,k}\in \{a_{2,k}\}_{k=1}^{\infty }\) is counted according to the minimum of all the multiplicities of \(a_{2,k}\) as the zero of \( f^n(z)\) and the \(m+1\) distinct entire functions in (3.31), respectively, such that:

$$\begin{aligned} \gamma _2(z)=z^{m_{2,0}}e^{\hat{g}(z)}\prod \limits _{k=1}^{\infty }\left( 1-\frac{z}{a_{2,k}}\right) e^{\frac{z}{a_{2,k}}+\frac{1}{2}\left( \frac{z}{a_{2,k}}\right) ^2 +\cdots +\frac{1}{m_{2,k}}\left( \frac{z}{a_{2,k}}\right) ^{m_{2,k}}}, \end{aligned}$$
(3.32)

where the product is taken over all \(a_{2,k}\ne 0.\) Here, \(m_{2,0}, m_{2,1}, \ldots , m_{2,n}, \ldots ,\) are certain nonnegative integers, and \(\hat{g}(z)\) is a finite constant or a nonconstant polynomial of degree less than or equal to \(q-1.\) Combining this with the assumption (H), we deduce:

$$\begin{aligned} N\left( r,\frac{1}{\gamma _2}\right) \le \sum \limits _{j=1}^m N\left( r,\frac{1}{H_j}\right) \le \sum \limits _{j=1}^m T\left( r, H_j\right) +O(1)\le O(r^{q-1})+O(\log r). \end{aligned}$$
(3.33)

By (3.32), (3.33) and Ash [1, Theorem 4.3.6], we have:

$$\begin{aligned} \lambda (\gamma _2)=\rho (\gamma _2)\le q-1. \end{aligned}$$
(3.34)

By dividing \(\gamma _2\) defined as in (3.32) on two sides of (3.9), we have:

$$\begin{aligned} \frac{f^n(z)}{\gamma _2(z)}=\sum \limits _{j=1}^{m}H_{2,j}(z)e^{\omega _{j}z^{q}}-\frac{p(z)f(z+\eta )}{\gamma _2(z)}, \end{aligned}$$
(3.35)

where:

$$\begin{aligned} H_{2,j}(z)=\frac{H_j(z)}{\gamma _2(z)} \quad \text {for}\quad 1\le j\le m. \end{aligned}$$
(3.36)

Then, by (3.35) and the supposition that the \(m+1\) distinct entire functions in (3.31) are linearly independent in the complex plane, we deduce that \(-\frac{p(z)f(z+\eta )}{\gamma _2(z)}\) and the m distinct entire functions \(H_{2,1}(z)e^{\omega _{1}z^{q}}, H_{2,2}(z)e^{\omega _{2}z^{q}}, \ldots , H_{2,m}(z)e^{\omega _{m}z^{q}}\) on the right-hand side of (3.35) are still linearly independent in the complex plane. Moreover, by (3.35) and the definition of the entire function \(\gamma _2\) as in (3.32), we can deduce that the entire function \(\frac{f^n(z)}{\gamma _2(z)}\) and the \(m+1\) entire functions \(H_{2,1}(z)e^{\omega _{1}z^{q}}, H_{2,2}(z)e^{\omega _{2}z^{q}}, \ldots , H_{2,m}(z)e^{\omega _{m}z^{q}}, -\frac{p(z)f(z+\eta )}{\gamma _2(z)}\) on the right-hand side of (3.35) have no common zeros in the complex plane. Furthermore, by (3.36) and the assumption (H), we deduce:

$$\begin{aligned} \frac{H_{2,i}(z)}{H_{2,j}(z)}e^{(\omega _{i}-\omega _{j})z^{q}}=\frac{H_i(z)}{H_j(z)}e^{(\omega _{i}-\omega _{j})z^{q}} \end{aligned}$$
(3.37)

is a transcendental meromorphic function for \(1\le i<j\le m.\) For convenience, now, we set:

$$\begin{aligned} \frac{f^n(z)}{\gamma _2(z)}= & {} g_{2,m+2}(z), \end{aligned}$$
(3.38)
$$\begin{aligned} -\frac{p(z)f(z+\eta )}{\gamma _2(z)}= & {} g_{2,m+1}(z) \end{aligned}$$
(3.39)

and

$$\begin{aligned} H_{2,j}(z)e^{\omega _{j}z^{q}}=g_{2,j}(z) \quad \text {for} \quad 1\le j\le m. \end{aligned}$$
(3.40)

Then, it follows by (3.38)–(3.40) that (3.35) can be rewritten as:

$$\begin{aligned} g_{2,m+2}=\sum \limits _{j=1}^{m+1} g_{2,j}, \end{aligned}$$
(3.41)

such that \(\max \nolimits _{1\le j\le m+1}\{|g_{2,j}(z)|\}>0\) for each \(z\in \mathbb {C},\) and \(g_{2,1}, g_{2,2}, \ldots , g_{2,m}, g_{2,m+1}\) on the right-hand side of (3.41) are \(m+1\) linearly independent transcendental entire functions. By geometry observation, we deduce from Lemma 2.5 and the supposition \(\rho _2(f)<1\) that:

$$\begin{aligned} N\left( r,\frac{1}{f(z+\eta )}\right) \le N\left( r+|\eta |,\frac{1}{f(z)}\right) \le N\left( r,\frac{1}{f(z)}\right) +o\left( \frac{T(r,f(z))}{r^{\delta }}\right) \end{aligned}$$
(3.42)

and

$$\begin{aligned} T\left( r,f(z+\eta )\right) \le T\left( r+|\eta |, f(z)\right) \le T(r,f(z))+o\left( \frac{T(r,f(z))}{r^{\delta }}\right) , \end{aligned}$$
(3.43)

as \(r\notin E_1\) and \(r\rightarrow \infty .\) Here, \(E_1\subset (0,+\infty )\) is a subset of finite logarithmic measure, while \(\delta \) is a fixed positive number satisfying \(\delta \in (0, 1-\rho _2)\) with \(\rho _2\) being a real number, such that \(\rho _2(f)=\limsup \nolimits _{r\rightarrow \infty }\frac{\log \log T(r,f)}{\log r}=\rho _2<1.\)

By (3.33), (3.37)–(3.42), Lemmas 2.2, 2.3, and Hayman [10, p.7], we deduce:

$$\begin{aligned} nN\left( r,\frac{1}{f(z)}\right)= & {} N\left( r,\frac{1}{g_{2,m+1}(z)}\right) +O(r^{q-1})\le T_2(r)+O(r^{q-1})\nonumber \\\le & {} \sum _{k=1}^{m+2}N_{m}\left( r,\frac{1}{g_{2,k}(z)}\right) +O(r^{q-1})+S_2(r)\nonumber \\\le & {} m\overline{N}\left( r,\frac{1}{f(z)}\right) +N\left( r,\frac{1}{f(z+\eta )}\right) +O(r^{q-1})+O(\log r)+S_2(r)\nonumber \\\le & {} mN\left( r,\frac{1}{f(z)}\right) +N\left( r,\frac{1}{f(z+\eta )}\right) +O(r^{q-1})+O(\log r)+S_2(r)\nonumber \\\le & {} mN\left( r,\frac{1}{f(z)}\right) +N\left( r+|\eta |,\frac{1}{f(z)}\right) +O(r^{q-1})+O(\log r)+S_2(r)\nonumber \\= & {} (m+1)N\left( r,\frac{1}{f(z)}\right) +o\left( \frac{T(r,f(z))}{r^{\delta }}\right) +O(r^{q-1})+O(\log r)+S_2(r),\nonumber \\ \end{aligned}$$
(3.44)

as \(r\not \in E_1\cup E\) and \(r\rightarrow \infty .\) Here:

$$\begin{aligned} T_2(r)=\frac{1}{2\pi }\int _0^{2\pi } u_2(re^{i\theta })\mathrm{{d}}\theta -u_2(0)\quad \text {with} \quad u_2(z)=\sup _{1\le k\le m+1}\{\log |g_{2,k}(z)|\}, \end{aligned}$$
(3.45)

and \(S_2(r)\) is a quantity satisfying:

$$\begin{aligned} S_2(r)=O(\log T_2(r))+O(\log r), \end{aligned}$$
(3.46)

as \(r\not \in E_1\cup E\) and \(r\rightarrow \infty ,\) where \(E\subset (0,+\infty )\) is a subset of finite linear measure.

By (3.44)–(3.46) and the assumption \(n\ge m+2\), we deduce:

$$\begin{aligned} N\left( r,\frac{1}{f(z)}\right) =O(r^{q-1})+O(\log T_2(r))+O(\log r)+S_2(r) \end{aligned}$$
(3.47)

and

$$\begin{aligned} T_2(r)\le (m+2)T(r,f)+O(r^{q-1})+O(\log r)+S_2(r), \end{aligned}$$
(3.48)

as \(r\not \in E_1\cup E\) and \(r\rightarrow +\infty .\)

Since f is a transcendental entire function, we deduce by (3.48) that:

$$\begin{aligned} S_2(r)=o(T(r,f))+O(r^{q-1}), \end{aligned}$$
(3.49)

as \(r\not \in E_1\cup E\) and \(r\rightarrow +\infty .\)

On the other hand, by rewriting (3.35), we have:

$$\begin{aligned} \frac{f^n(z)}{\gamma _2(z)}+\frac{p(z)f(z+\eta )}{\gamma _2(z)} =\sum \limits _{j=1}^{m}H_{2,j}(z)e^{\omega _{j}z^{q}}, \end{aligned}$$
(3.50)

By (3.33), (3.34), (3.43) and Definition 1.1, we deduce:

$$\begin{aligned} N\left( r, \frac{1}{\frac{f^n(z)}{\gamma _2(z)}+\frac{p(z)f(z+\eta )}{\gamma _2(z)}}\right)\le & {} T\left( r, \frac{f^n(z)}{\gamma _2(z)}+\frac{p(z)f(z+\eta )}{\gamma _2(z)}\right) +O(1)\nonumber \\\le & {} T(r,f^n(z))+T(r,f(z+\eta ))+O(r^{q-1})+O(1)\nonumber \\\le & {} T(r,f^n(z))+T(r,f(z+\eta ))+O(r^{q-1})+O(1)\nonumber \\\le & {} (n+1)T(r,f(z))+O(r^{q-1})+o\left( \frac{T(r,f(z))}{r^{\delta }}\right) ,\nonumber \\ \end{aligned}$$
(3.51)

as \(r\notin E_1\) and \(r\rightarrow \infty .\) Here, \(\delta \) is the fixed positive number defined as in (3.43).

On the other hand, by (3.33), (3.34), (3.36), (3.50), Lemma 2.4, and the assumption (H), we deduce:

$$\begin{aligned} N\left( r, \frac{1}{\frac{f^n(z)}{\gamma _2(z)}+\frac{p(z)f(z+\eta )}{\gamma _2(z)}}\right)= & {} N\left( r, \frac{1}{\sum \limits _{j=1}^{m}H_{2,j}(z)e^{\omega _{j}z^{q}}}\right) \nonumber \\= & {} N\left( r, \frac{1}{H_{2,1}(z)+\sum \limits _{j=2}^{m}H_{2,j}(z)e^{(\omega _{j}-\omega _{1})z^{q}}}\right) \nonumber \\= & {} C(co(\hat{W}_0))\frac{r^q}{2\pi }+o(r^q), \end{aligned}$$
(3.52)

where \(\hat{W}_0\) is defined as in (3.29).

By (3.51) and (3.52), we deduce:

$$\begin{aligned} C(co(\hat{W}_0))\frac{r^q}{2\pi }\le (n+1)T(r,f(z))+o\left( \frac{T(r,f(z))}{r^{\delta }}\right) +o(r^q), \end{aligned}$$
(3.53)

as \(r\notin E_1\) and \(r\rightarrow \infty .\) Here, \(\delta \) is the fixed positive number defined as in (3.43).

By dividing \(f(z+\eta )\) on two sides of (3.35), we have:

$$\begin{aligned} \frac{f^n(z)}{\gamma _2(z)f(z+\eta )}=\sum \limits _{j=1}^{m}\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )} -\frac{p(z)}{\gamma _2(z)}. \end{aligned}$$
(3.54)

By (3.33), (3.34), (3.36), (3.43), (3.53), (3.54), Lemma 2.4, and the assumption \(n\ge m+2\), we deduce:

$$\begin{aligned} T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right)\ge & {} T\left( r,f^n(z)\right) -T(r,\gamma _2(z)f(z+\eta ))\nonumber \\\ge & {} T(r,f^n(z))-T(r,f(z+\eta ))-T(r,\gamma _2(z))\nonumber \\\ge & {} T(r,f^n(z))-T(r,f(z))-T(r,\gamma _2(z))\nonumber \\&+o\left( \frac{T(r,f(z))}{r^{\delta }}\right) \nonumber \\\ge & {} (n-1)T(r,f(z))+O(r^{q-1})+o\left( \frac{T(r,f(z))}{r^{\delta }}\right) \nonumber \\\ge & {} \frac{n-1}{n+1}\cdot \frac{C(co(\hat{W}_0))}{2\pi }r^q+o(r^q) \end{aligned}$$
(3.55)

and

$$\begin{aligned}&T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) \nonumber \\&\quad \le T\left( r,\sum \limits _{j=1}^{m}H_{2,j}(z)e^{\omega _{j}z^{q}}\right) +T\left( r,\frac{1}{f(z+\eta )}\right) +T\left( r,-\frac{p(z)}{\gamma _2(z)}\right) \nonumber \\&\quad \le C(co(\hat{W}_0))\frac{r^q}{2\pi }+ T(r,f(z))+o\left( \frac{T(r,f(z))}{r^{\delta }}\right) +O(r^{q-1})+O(\log r)\nonumber \\&\quad \le C(co(\hat{W}_0))\frac{r^q}{2\pi }+\frac{1}{n-1}T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) +o(r^q), \end{aligned}$$
(3.56)

as \(r\notin E_1\) and \(r\rightarrow \infty .\)

By (3.55) and (3.56), we deduce:

$$\begin{aligned} \frac{n-1}{n+1}\frac{C(co(\hat{W}_0))}{2\pi }r^q+o(r^q)\le T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) \le \frac{n-1}{n-2} \frac{C(co(\hat{W}_0))}{2\pi }r^q+o(r^q) \end{aligned}$$
(3.57)

and

$$\begin{aligned} \frac{1}{n+1}\cdot \frac{C(co(\hat{W}_0))}{2\pi }r^q+o(r^q) \le T\left( r,f\right) \le \frac{1}{n-2}\cdot \frac{C(co(\hat{W}_0))}{2\pi }r^q+o(r^q), \end{aligned}$$
(3.58)

as \(r\rightarrow \infty \) and \(r\notin E_1.\)

By (3.54). we consider the following two subcases.

Subcase 2.2.1.2.1.   Suppose that there exists some subset \(I\subset (0,+\infty )\) of infinite logarithmic measure, i.e., \(\int _{I\cap [1,+\infty )}\frac{\mathrm{{d}}t}{t}=+\infty ,\) and there exists one and only one of the m distinct meromorphic functions \(\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}, \frac{H_{2,2}(z)e^{\omega _{2}z^{q}}}{f(z+\eta )},\ldots , \frac{H_{2,m}(z)e^{\omega _{m}z^{q}}}{f(z+\eta )}\) in (3.54), say \(\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )},\) such that:

$$\begin{aligned} T\left( r,\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}\right) \le A_1 r^{q-1+\varepsilon }, \end{aligned}$$
(3.59)

as \(r\in {I}\) and \(r\rightarrow +\infty ,\) while:

$$\begin{aligned} T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \ge A_j r^{q} \quad \text {for}\quad 2\le j\le m, \end{aligned}$$

as \(r\in {I}\) and \(r\rightarrow +\infty .\) Here, \(A_j>0\) is a constant for \(1\le j\le m.\) This together with (3.43), the assumption (H), and Hayman [10, p.7], gives:

$$\begin{aligned} A_j r^{q}\le T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \le \tilde{A}_j r^{q} \quad \text {for}\quad 2\le j\le m, \end{aligned}$$
(3.60)

as \(r\in {I}\) and \(r\rightarrow +\infty .\) Here, \( \tilde{A}_j\) is a positive number ,such that \(A_j<\tilde{A}_j\) for \(2\le j\le m.\)

By (3.33), (3.34), (3.42), (3.46)–(3.49), and (3.58), we deduce:

$$\begin{aligned} T\left( r,-\frac{p(z)}{\gamma _2(z)}\right) + N\left( r,\frac{1}{f(z+\eta )}\right) +N\left( r,\frac{1}{f(z)}\right) =o(T(r,f(z))), \end{aligned}$$
(3.61)

as \(r\not \in E_1\cup E\) and \(r \rightarrow +\infty .\)

By (3.54), (3.57), (3.59), (3.61), and Lemma 2.1, we deduce that the positive integer m satisfies \(m\ge 2.\)

By (3.57), (3.59), (3.60), and the assumption \(n\ge m+2\), we deduce:

$$\begin{aligned} T\left( r,\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}\right) =o\left( T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) \right) \end{aligned}$$
(3.62)

and

$$\begin{aligned} T\left( r,\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}\right) =o\left( T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \right) \quad \text {for}\quad 2\le j\le m, \end{aligned}$$
(3.63)

as \(r\in {I}\) and \(r\rightarrow +\infty .\)

By (3.57), (3.60), (3.61), and the assumption \(n\ge m+2\), we deduce:

$$\begin{aligned} T\left( r,-\frac{p(z)}{\gamma _2(z)}\right) =o\left( T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) \right) \end{aligned}$$
(3.64)

and

$$\begin{aligned} T\left( r,-\frac{p(z)}{\gamma _2(z)}\right) =o\left( T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \right) \quad \text {for}\quad 2\le j\le m, \end{aligned}$$
(3.65)

as \(r\in {I}\) and \(r\rightarrow +\infty .\)

By (3.33), (3.42), (3.47), (3.57), (3.58), (3.60), the assumption (H), and Hayman [10, p.7], we deduce:

$$\begin{aligned}&N\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) +N\left( r,\frac{\gamma _2(z)f(z+\eta )}{f^n(z)}\right) +\sum \limits _{j=2}^{m}N\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \nonumber \\&\quad +\sum \limits _{j=2}^{m}N\left( r,\frac{f(z+\eta )}{H_{2,j}(z)e^{\omega _{j}z^{q}}}\right) =o\left( T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) \right) \end{aligned}$$
(3.66)

and

$$\begin{aligned}&N\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) +N\left( r,\frac{\gamma _2(z)f(z+\eta )}{f^n(z)}\right) +\sum \limits _{j=2}^{m}N\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \nonumber \\&\quad +\sum \limits _{j=2}^{m}N\left( r,\frac{f(z+\eta )}{H_{2,j}(z)e^{\omega _{j}z^{q}}}\right) =o\left( T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \right) , \end{aligned}$$
(3.67)

as \(r\in {I}\) and \(r\rightarrow +\infty .\)

By (3.54), (3.62)-(3.67), and Lemma 2.1, we deduce:

$$\begin{aligned} \frac{H_{2,1}(z)e^{\omega _jz^{q}}}{f(z+\eta )}-\frac{p(z)}{\gamma _2(z)}\equiv 0 \end{aligned}$$

and

$$\begin{aligned} \frac{f^n(z)}{\gamma _2(z)f(z+\eta )}=\sum \limits _{j=2}^{m}\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}, \end{aligned}$$

and so, it follows by (3.36) that:

$$\begin{aligned} f(z+\eta )= \frac{\gamma _2(z)H_{2,1}(z)e^{\omega _{1}z^{q}}}{p(z)}= \frac{H_1(z)e^{\omega _{1}z^{q}}}{p(z)} \end{aligned}$$
(3.68)

and

$$\begin{aligned} f^n(z+\eta )=\gamma _2(z+\eta )\sum \limits _{j=2}^{m}H_{2,j}(z+\eta )e^{\omega _{j}(z+\eta )^{q}}=\sum \limits _{j=2}^{m}H_j(z+\eta )e^{\omega _{j}(z+\eta )^{q}}. \end{aligned}$$
(3.69)

By substituting (3.68) into (3.69), we deduce:

$$\begin{aligned} H^n_1(z)e^{n\omega _{1}z^{q}}= p^n(z)\sum \limits _{j=2}^{m}L_j(z)e^{\omega _j z^q}, \end{aligned}$$
(3.70)

where

$$\begin{aligned} L_j(z)=H_j(z+\eta )e^{\omega _jP_{q-1}(z)}\quad \text {for} \quad 2\le j \le m. \end{aligned}$$
(3.71)

Here:

$$\begin{aligned} P_{q-1}(z)=\sum \limits _{k=1}^q\left( {\begin{array}{c}q\\ k\end{array}}\right) \eta ^kz^{q-k}. \end{aligned}$$
(3.72)

By (3.71), (3.72), Lemma 2.6, and Hayman [10, p.7], we deduce:

$$\begin{aligned} T(r,L_j(z))\le & {} T\left( r, H_j(z+\eta )\right) +T\left( r,e^{\omega _jP_{q-1}(z)}\right) \nonumber \\= & {} T\left( r, H_j(z)\right) +T\left( r,e^{\omega _jP_{q-1}(z)}\right) +O(r^{q-2+\varepsilon })+O(\log r)\nonumber \\\le & {} T\left( r,e^{\omega _jP_{q-1}(z)}\right) +O(r^{q-2+\varepsilon })+O(r^{q-1})+O(\log r)\nonumber \\= & {} \frac{q|\eta |r^{q-1}}{\pi }\left( 1+o(1)\right) +O(r^{q-2+\varepsilon })+O(r^{q-1})+O(\log r)=o(r^q)\nonumber \\ \end{aligned}$$
(3.73)

for \(2\le j\le m,\) as \(r\rightarrow +\infty .\)

We consider the following two subcases.

Subcase 2.2.1.2.1.1. Suppose that \(m=2.\) Then, (3.70) can be rewritten as:

$$\begin{aligned} e^{(n\omega _{1}-\omega _2)z^{q}}=\frac{p^n(z)L_2(z)}{H^n_1(z)}. \end{aligned}$$
(3.74)

By (3.71), (3.73), (3.74), the assumption (H), and Hayman [10, p.7], we deduce:

$$\begin{aligned} T\left( r, \frac{p^nL_2}{H^n_1}\right) =O(r^{q-1})+O(\log r) \end{aligned}$$
(3.75)

and

$$\begin{aligned} T(r,H^n_1)+T(r,p^nL_j)=O(r^{q-1})+O(\log r)\quad \text {for}\quad 2\le j\le m. \end{aligned}$$
(3.76)

By (3.74), (3.75) and Hayman [10, p.7], we deduce:

$$\begin{aligned} \frac{|n\omega _{1}-\omega _2|}{\pi }r^q(1+o(1))&=T\left( r, e^{(n\omega _{1}-\omega _2)z^{q}}\right) =T\left( r, \frac{p^n(z)L_2(z)}{H^n_1(z)}\right) \\&=O(r^{q-1})+O(\log r), \end{aligned}$$

which implies that \(\omega _2=n\omega _{1}.\) This together with (3.70)-(3.72) implies that:

$$\begin{aligned} H^n_1(z)=p^n(z)H_2(z+\eta )e^{\omega _2P_{q-1}(z)}, \end{aligned}$$
(3.77)

where \(P_{q-1}(z)\) is defined as in (3.72).

By (3.68), (3.72), (3.77), and \(\omega _2=n\omega _{1}\), we get the conclusion (i) of Theorem 1.8.

Subcase 2.2.1.2.1.2. Suppose that \(m\ge 3.\) Then, by (3.70), (3.76), Lemma 2.4 and the fact \(p^nL_j\not \equiv 0\) for \(2\le j\le m\), we deduce:

$$\begin{aligned} C(co(\tilde{W_0}))\frac{r^q}{2\pi }+o(r^q)&=N\left( r,\frac{1}{p^n(z)L_2(z)+p^n(z)\sum \limits _{j=3}^{m}L_j(z)e^{(\omega _j-\omega _2) z^q}}\right) \\&=N\left( r,\frac{1}{p^n(z)\sum \limits _{j=2}^{m}L_j(z)e^{\omega _j z^q}}\right) =N\left( r,\frac{1}{H^n_1(z)e^{n\omega _{1}z^{q}}}\right) \\&=N\left( r,\frac{1}{H^n_1(z)}\right) \\&\le T(r,H^n_1(z))+O(1)=O(r^{q-1})+O(\log r)+O(1)\\&=o(r^q), \end{aligned}$$

which is impossible. Here, \(\tilde{W}_0 = \{0, \overline{w_3-w_2}, \overline{w_4-w_2}, \ldots , \overline{w_m-w_2}\}.\)

Subcase 2.2.1.2.2.   Suppose that there exists some subset \(I\subset (0,+\infty )\) with infinite logarithmic measure, i.e., \(\int _{I\cap [1,+\infty )}\frac{\mathrm{{d}}t}{t}=+\infty ,\) and there exist at least two of the m distinct meromorphic functions \(\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}, \) \(\frac{H_{2,2}(z)e^{\omega _{2}z^{q}}}{f(z+\eta )},\) \(\ldots ,\) \(\frac{H_{2,m}(z)e^{\omega _{m}z^{q}}}{f(z+\eta )}\) in (3.54), say \(\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}\) and \(\frac{H_{2,2}(z)e^{\omega _{2}z^{q}}}{f(z+\eta )},\) such that (3.59) holds, such that (3.60) holds for \(3\le j\le m,\) and such that:

$$\begin{aligned} T\left( r,\frac{H_{2,2}(z)e^{\omega _{2}z^{q}}}{f(z+\eta )}\right) \le B_1 r^{q-1+\varepsilon }, \end{aligned}$$
(3.78)

as \(r\in {I}\) and \(r\rightarrow +\infty .\) Here, \(B_1>0\) is a constant.

By (3.36), (3.59), (3.78), the assumption (H), and Hayman [10, p.7], we deduce:

$$\begin{aligned} \frac{|\omega _{1}-\omega _{2})|}{\pi }r^{q}\left( 1+o(1)\right)&=T\left( r,\frac{H_1(z)}{H_2(z)}e^{(\omega _{1}-\omega _{2})z^{q}}\right) +O(r^{q-1})+O(\log r)\\&=T\left( r,\frac{\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}}{\frac{H_{2,2}(z)e^{\omega _{2}z^{q}}}{f(z+\eta )}}\right) +O(r^{q-1})+O(\log r)\\&\le T\left( r,\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}\right) +T\left( r,\frac{H_{2,2}(z)e^{\omega _{2}z^{q}}}{f(z+\eta )}\right) \\&\quad +O(r^{q-1})+O(\log r)+O(1)\\&\le A_1 r^{q-1+\varepsilon }+B_1 r^{q-1+\varepsilon }+O(r^{q-1})+O(\log r)+O(1)\\&\quad =o(r^q), \end{aligned}$$

as \(r\in {I}\) and \(r\rightarrow +\infty .\) This is impossible.

Subcase 2.2.1.2.3.   Suppose that the m distinct meromorphic functions \(\frac{H_{2,1}(z)e^{\omega _{1}z^{q}}}{f(z+\eta )}, \) \(\frac{H_{2,2}(z)e^{\omega _{2}z^{q}}}{f(z+\eta )},\) \(\ldots ,\) \(\frac{H_{2,m}(z)e^{\omega _{m}z^{q}}}{f(z+\eta )}\) in (3.54) are such that:

$$\begin{aligned} T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \ge C_j r^{q} \quad \text {for}\quad 1\le j\le m, \end{aligned}$$
(3.79)

as \(r\not \in E_1\) and \(r\rightarrow +\infty .\) Here, \(C_j>0\) is a constant for \(1\le j\le m.\)

On the other hand, by (3.34), (3.36), (3.43), (3.58), the assumption (H), and Hayman [10, p.7], we deduce:

$$\begin{aligned} T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right)\le & {} T\left( r,H_{2,j}(z)e^{\omega _{j}z^{q}}\right) +T\left( r,\frac{1}{f(z+\eta )}\right) \nonumber \\\le & {} \frac{|\omega _{j}|}{\pi }r^{q}\left( 1+o(1)\right) +T(r,f(z+\eta ))+O(r^{q-1})+O(\log r)\nonumber \\\le & {} \frac{|\omega _{j}|}{\pi }r^{q}\left( 1+o(1)\right) +T(r,f(z))+o\left( \frac{T(r,f(z))}{r^{\delta }}\right) \nonumber \\&+O(r^{q-1})+O(\log r)\nonumber \\\le & {} \left( \frac{|\omega _{j}|}{\pi }+\frac{1}{n-2}\cdot \frac{C(co(\hat{W}_0))}{2\pi }\right) r^{q}\left( 1+o(1)\right) +o(r^q)\nonumber \\ \end{aligned}$$
(3.80)

for \(1\le j\le m,\) as \(r\not \in E_1\) and \(r\rightarrow +\infty .\)

By (3.79) and (3.80), we deduce:

$$\begin{aligned} C_j r^{q}\le T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \le \left( \frac{|\omega _{j}|}{\pi }+\frac{1}{n-2}\cdot \frac{C(co(\hat{W}_0))}{2\pi }\right) r^{q}\left( 1+o(1)\right) +o(r^q) \end{aligned}$$
(3.81)

for \(1\le j\le m,\) as \(r\not \in E_1\) and \(r\rightarrow +\infty .\)

Next, in the same manner as in Subcase 2.2.1.2.1, we can deduce by (3.57) and (3.81) that:

$$\begin{aligned}&T\left( r,-\frac{p(z)}{\gamma _2(z)}\right) =o\left( T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) \right) , \end{aligned}$$
(3.82)
$$\begin{aligned}&T\left( r,-\frac{p(z)}{\gamma _2(z)}\right) =o\left( T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \right) \quad \text {for}\quad 1\le j\le m, \end{aligned}$$
(3.83)
$$\begin{aligned}&N\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) +N\left( r,\frac{\gamma _2(z)f(z+\eta )}{f^n(z)}\right) +\sum \limits _{k=1}^{m}N\left( r,\frac{H_{2,k}(z)e^{\omega _{k}z^{q}}}{f(z+\eta )}\right) \nonumber \\&\quad +\sum \limits _{k=1}^{m}N\left( r,\frac{f(z+\eta )}{H_{2,k}(z)e^{\omega _{k}z^{q}}}\right) =o\left( T\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) \right) \end{aligned}$$
(3.84)

and

$$\begin{aligned}&N\left( r,\frac{f^n(z)}{\gamma _2(z)f(z+\eta )}\right) +N\left( r,\frac{\gamma _2(z)f(z+\eta )}{f^n(z)}\right) +\sum \limits _{k=1}^{m}N\left( r,\frac{H_{2,k}(z)e^{\omega _{k}z^{q}}}{f(z+\eta )}\right) \nonumber \\&\quad +\sum \limits _{k=1}^{m}N\left( r,\frac{f(z+\eta )}{H_{2,k}(z)e^{\omega _{k}z^{q}}}\right) =o\left( T\left( r,\frac{H_{2,j}(z)e^{\omega _{j}z^{q}}}{f(z+\eta )}\right) \right) \quad \text {for}\quad 1\le j\le m,\nonumber \\ \end{aligned}$$
(3.85)

as \(r\not \in E_1\) and \(r\rightarrow +\infty .\)

By (3.54), (3.82)–(3.85), Lemma 2.1, and the supposition that the \(m+1\) entire functions in (3.31) are linearly independent in the complex plane, we deduce that \(-\frac{p}{\gamma _2}=0,\) which is impossible.

Suppose that \( f^n(z)\) and the \(m+1\) distinct entire functions in (3.31) have at most finitely many distinct nonzero common zeros in the complex plane, say \(a_{2,1}, a_{2,2}, \ldots , a_{2,l_2}\) are all the distinct nonzero common zeros of \( f^n(z)\) and the \(m+1\) distinct entire functions of (3.31) in the complex plane, where \(l_2\) is a positive integer, and each such common zero \(a_{2,j}\in \left\{ a_{2,j}\right\} _{j=1}^{l_2}\) is counted according to the minimum positive integer \(\tilde{m}_{2,j}\) of all the multiplicities of \(a_{2,j}\) as the zero of \( f^n(z)\) and the \(m+1\) distinct entire functions of (3.31), respectively. Now, we let:

$$\begin{aligned} P_{1,2}(z)=z^{m_{2,0}}\prod \limits _{j=1}^{l_2}\left( z-a_{2,j}\right) ^{\tilde{m}_{2,j}}, \end{aligned}$$
(3.86)

where \(m_{2,0}\) is the nonnegative integer that is defined as in (3.32). Next, we replace the polynomial \(P_{1,2}\) defined as in (3.86) with the entire function \(\gamma _2\) defined as in (3.32), and then use the lines as the above, we can get a contradiction.

Subcase 2.2.2.   Suppose that \(H_0\not \equiv 0.\) Then, (1.1) can be rewritten as:

$$\begin{aligned} f^n(z)=H_0(z)+H_1(z)e^{\omega _{1}z^{q}}+H_2(z)e^{\omega _{2}z^{q}}+\cdots +H_m(z)e^{\omega _{m}z^{q}}-p(z)f(z+\eta ). \end{aligned}$$
(3.87)

By (3.87), we deduce that either that the \(m+2\) entire functions:

$$\begin{aligned}&H_0(z), H_1(z)e^{\omega _{1}z^{q}}, H_2(z)e^{\omega _{2}z^{q}}, \ldots , H_{m-1}(z)e^{\omega _{(m-1)}z^{q}}, H_{m}(z)e^{\omega _{(m)}z^{q}},\nonumber \\&\quad -p(z)f(z+\eta ) \end{aligned}$$
(3.88)

are \(m+2\) linearly dependent entire functions in the complex plane, or that the \(m+2\) entire functions in (3.88) are linearly independent entire functions in the complex plane.

Next, in the same manner as in the proof of Subcase 2.2.1.1 and Subcase 2.2.1.2, by Lemmas 2.5 and 2.6, we deduce (3.42) and (3.43). By (3.42), (3.43), (3.87), Lemma 2.4, and the supposition \(H_0\not \equiv 0\), we deduce \(n= m+2,\)

$$\begin{aligned} \frac{C(co(W_0))}{2(n+1)\pi }r^q+o(r^q)\le T(r,f(z))\le \frac{C(co(W_0))}{2(n-1)\pi }r^q+o(r^q) \end{aligned}$$
(3.89)

and

$$\begin{aligned} B_2T(r,f(z))\le&{} N\left( r,\frac{1}{f(z)}\right) \le T(r,f)+O(1), \quad \text{ as }\quad r\not \in E_1\cup E\quad \text{ and }\nonumber \\ \quad r\rightarrow +\infty . \end{aligned}$$
(3.90)

Here \(B_2\) is some positive constant.

By Definition 1.1 and the standard reasoning of removing an exceptional set (cf. [12, Lemma 1.1.2]). we deduce by (3.89) and (3.90) that:

$$\begin{aligned} \lambda (f)=\rho (f)=q. \end{aligned}$$
(3.91)

By (3.91) and the obtained result \(n=m+2\), we get the conclusion (ii) of Theorem 1.8. This completes the proof of Theorem 1.8.

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Acknowledgements

The authors wish to express their thanks to the referee for his/her valuable suggestions and comments. The first author of this paper also wants to express his sincere thanks to Professor R. Korhonen, Professor J. Heittokangas, and Professor I. Laine for their help and guidance during his visit at the Department of Physics and Mathematics, University of Eastern Finland from June 10, 2019 to August 20, 2019. This sped up the completion of this article.

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Project was supported in part by the NSFC (no. 11171184), the NSF of Shandong Province, China (no. ZR2019MA029), and the FRFCU (no. 3016000841964007).

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Li, XM., Hao, CS. & Yi, HX. On the Growth of Meromorphic Solutions of Certain Nonlinear Difference Equations. Mediterr. J. Math. 18, 56 (2021). https://doi.org/10.1007/s00009-020-01696-z

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Keywords

  • Nevanlinna’s theory
  • Cartan’s version of Nevanlinna theory
  • difference Nevanlinna’s theory
  • meromorphic functions
  • nonlinear difference equations

Mathematics Subject Classification

  • Primary 30D35
  • Secondary 39B32